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Math Help - help with maximum curvature

  1. #1
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    help with maximum curvature

    What is the maximal curvature of the curve y = ln cos x?
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  2. #2
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    Find an expression for the curvature using

    Then find the maximum.
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  3. #3
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    I'm still having

    problems.
    y' = -sinx/cosx or -tanx
    y'' = -secx^2
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  4. #4
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    Put those in to get

    \kappa=\frac{\sec^2 x}{(1+\tan^2 x)^{3/2}}

    Use 1+\tan^2 x=\sec^2 x

    and

    \kappa=\frac{\sec^2 x}{\sec^3 x}.

    So

    \kappa=\frac{1}{\sec x }=\cos x.
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